Consider a quantum particle moving in the $x$- direction and hitting a wall of higher energy than it has. Classically, the particle cannot pass through the wall. Quantum mechanically, it can.

Let $u(x)$ be a solution to the Schrödinger equation when the particle is in the region where the potential barrier is nonzero.

$$u(x) = A e^{(-lx)}+B e^{(lx)}$$

Why is it that u(x) also cannot be normalized (the integral over all space of the $|u(x)|^2$ is infinity) and what is the physical implication of not being normalizable while it is tunneling through this barrier?

I also read that in order to be an acceptable solution of the Schrödinger equation, the solution must be continuous and must be normalizable).

  • $\begingroup$ That solutions must be normalizable is not true, see free particle. They shd be that they can define dustributions. $\endgroup$ – lalala Mar 30 '18 at 15:18

The barrier, in principle, isn’t infinite in length. Thus, the contribution to the normalization factor in that region is given by


which is finite so long as the barrier is finite in width. If instead the barrier, for example, ran from $x=0$ to $x\to\infty$, then naturally this wouldn’t be the case unless $B=0$.

In summary, you don’t need to integrate over all of space, since the solution you wrote down only applies to a small region of space.


Your solution $u(x)$, only holds within the barrier. In addition, it is always a decaying exponential with respect to which direction the particle is coming from.

If we look at the finite well, the eigenstates are normalizable because the exponential solution only exists in the classically forbidden region. In the allowed region, the eigenstates are sinusoidal. The solutions are combined piecewise to obtain the entire normalizable wave-function of the finite square well. $$ \psi = \Bigg\{\begin{smallmatrix} Ae^{-k x} & x <-L/2\\ Be^{-ik x} + B^*e^{-ik x} & -L/2<x<L/2\\ Ce^{-k x} & x > L/2 \end{smallmatrix} $$

Something to remember is that any wave-function can be constructed as a linear superposition of the eigenstates. You can always construct a wave-function with a finite probability distribution.


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